The expression is evaluated by applying Euler's formula:
#color(red)(e^(ix)=cosx+isinx)#
We will use the following trigonometric identities:
#color(blue)(cos(2pi+alpha)=cosalpha#
#color(blue)(cos(-alpha)=cosalpha)#
#color(blue)(sin(-alpha)=-sinalpha)#
Let us compute #e^(pi/2i)and e^((5pi)/3i)# by applying the above formula
#e^(pi/2i)=cos(pi/2)+isin(pi/2)#
#rArre^(pi/2i)=0+i(1)#
#rArre^(pi/2i)=i#
#e^((5pi)/3i)=cos((5pi)/3)+isin((5pi)/3)#
#rArre^((5pi)/3i)=cos((6pi)/3-pi/3)+isin((6pi)/3-pi/3)#
#rArre^((5pi)/3i)=cos(2pi-pi/3)+isin(2pi-pi/3)#
#rArre^((5pi)/3i)=color(blue)(cos(-pi/3)+isin(-pi/3))#
#rArre^((5pi)/3i)=color(blue)(cos(pi/3)-isin(pi/3)#
#rArre^((5pi)/3i)=1/2-isqrt3/2#
#e^(pi/2i)-e^((5pi)/3i)#
#=i-(1/2-isqrt3/2)#
#=i-1/2+isqrt3/2#
#=-1/2+i(1+sqrt3/2)#
Therefore,
#e^(pi/2i)-e^((5pi)/3i)=-1/2+i(1+sqrt3/2)#
